IP AMaths Notes (Upper Sec, Year 3-4): 16) Applications of Differentiation

Study guide08 Nov 2025, 00:00 ZUpdated 17 Jul 2026

Tangents, normals, stationary points, and related rates in IP AMaths.

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For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.

How this chapter applies

Eclat core: increasing and decreasing functions, stationary points including stationary inflexion, second-derivative classification, tangents, normals, connected rates, and optimisation form the main route.
School-sensitive extension: unfamiliar related-rate geometry and multi-constraint optimisation may vary by school.
2027 national comparison: K341 Topic C1 includes monotonicity, stationary points, the second-derivative test, gradients, tangents, normals, connected rates, maxima, and minima.
Check your school: confirm sign-table, second-derivative, and justification conventions.
Exam-track route: use the separate O-Level and SEC G3 Additional Mathematics notes for K341 topic ownership.

Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 16) Applications of Differentiation cover?
A: Tangents, normals, stationary points, and related rates in IP AMaths.

Apply derivatives to describe gradient, optimise functions, and connect rates.

Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.

New to the Integrated Programme? Start with What is IP? | Browse all free IP notes.

The core idea is simple: Applications of differentiation turn gradients into decisions.

Use it as a working check: Use derivatives to find tangent gradients, normal gradients, stationary points, and related rates. Always classify stationary points before naming a maximum or minimum.

Then go one layer deeper: Example: if y' = 0 at x = 3, test the second derivative or a sign chart before calling it a turning point. If y'' is positive, it is a local minimum.

Choosing the application route

Start by deciding what the derivative represents in the question. A derivative can be a gradient, a condition for a stationary point, or a link between changing quantities.

Question cue

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Marcus Pang·Managing Director (Maths)

Step	Tangent line	Normal line	Common trap
1. Differentiate	Find $y'$ .	Find $y'$ first, because the normal depends on the tangent gradient.	Starting with the reciprocal before finding the tangent gradient.
2. Substitute the contact value	Put $x = a$ into $y'$ to get $m_\text{tangent}$ .	Use the same contact point as the tangent.	Using a nearby point from the sketch instead of the exact point on the curve.
3. Find the point	Put $x = a$ into the original curve to get $(a, y(a))$ .	The normal also passes through $(a, y(a))$ .	Substituting
4. Write the line	Use $y - y_1 = m(x - x_1)$ .	Use $m_\text{normal} = -1/m_\text{tangent}$

Setup question	What to write	Common trap
Which quantities are linked geometrically?	An equation such as area, volume, or Pythagoras that connects them.	Differentiating before removing unnecessary variables.
Which rate is given?	Mark it as a derivative with respect to time, such as $\dfrac{\mathrm{d}V}{\mathrm{d}t}$ .	Treating $\dfrac{\mathrm{d}V}{\mathrm{d}t}$ as if it were just $V$ .
Which rate is required?	Rearrange after differentiating to isolate the requested derivative.	Giving $\dfrac{\mathrm{d}V}{\mathrm{d}t}$ when the question asks for $\dfrac{\mathrm{d}h}{\mathrm{d}t}$
At what instant is the rate needed?	Substitute the instant-specific length, radius, height, or volume after differentiating.	Substituting changing values too early and turning variables into constants.

Evidence	Classification	What to write
$y'$ changes from positive to negative.	Local maximum.	The curve rises before the point and falls after it.
$y'$ changes from negative to positive.	Local minimum.	The curve falls before the point and rises after it.
$y'$ keeps the same sign on both sides.	Stationary point of inflexion.	The curve flattens but does not turn.
$y'' > 0$ at the stationary point.	Local minimum.	The curve is concave upward there.
$y'' < 0$ at the stationary point.	Local maximum.	The curve is concave downward there.
$y'' = 0$ at the stationary point.	Test is inconclusive.	Use a sign chart for $y'$ instead of guessing.

IP AMaths Notes (Upper Sec, Year 3-4): 16) Applications of Differentiation

How this chapter applies

Choosing the application route

Sources

1 Common tasks

Tangent and normal line checkpoint

Related-rates setup checkpoint

2 Worked example - Tangent and normal

3 Worked example - Optimisation

Stationary-point classification checkpoint

4 Worked example - Classifying stationary points

5 Practice Quiz

6 Try this

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